# A Group Homomorphism is Injective if and only if the Kernel is Trivial ## Problem 144

Let $G$ and $H$ be groups and let $f:G \to K$ be a group homomorphism. Prove that the homomorphism $f$ is injective if and only if the kernel is trivial, that is, $\ker(f)=\{e\}$, where $e$ is the identity element of $G$. Add to solve later

## Definitions/Hint.

We recall several relevant definitions.

• A group homomorphism $f:G\to H$ is a map such that for any $g_1, g_2 \in G$, we have
$f(g_1g_2)=f(g_1)f(g_2).$
• A group homomorphism $f:G \to H$ is injective if for any $g_1, g_2 \in G$
the equality
$f(g_1)=f(g_2)$ implies $g_1=g_2$.
• The kernel of a group homomorphism $f:G \to H$ is a set of all elements of $G$ that is mapped to the identity element of $H$.
Namely,
$\ker(f)=\{g\in G \mid f(g)=e’\},$ where $e’$ is the identity element of $H$.

## Proof.

### Injective $\implies$ the kernel is trivial

Suppose the homomorphism $f: G \to H$ is injective.
Then since $f$ is a group homomorphism, the identity element $e$ of $G$ is mapped to the identity element $e’$ of $H$. Namely, we have $f(e)=e’$.
If $g \in \ker(f)$, then we have $f(g)=e’$, and thus we have
$f(g)=f(e).$ Since $f$ is injective, we must have $g=e$. Thus we have $\ker(f)=\{e\}$.

### The kernel is trivial $\implies$ injective

On the other hand, suppose that $\ker(f)=\{e\}$.
If $g_1, g_2$ are elements of $G$ such that
$f(g_1)=f(g_2), \tag{*}$ then we have
\begin{align*}
f(g_1g_2^{-1})&=f(g_1)f(g_2^{-1}) \quad \text{ since } f \text{ is a homomorphism}\\
&=f(g_1)f(g_2)^{-1} \quad \text{ since } f \text{ is a homomorphism}\\
&=f(g_1)f(g_1)^{-1} \quad \text{ by } (*)\\
&=e’.
\end{align*}

In the second step, we used the fact $f(g_2^{-1})=f(g_2)^{-1}$, which is proved in the post “Group Homomorphism Sends the Inverse Element to the Inverse Element“.

Thus the element $g_1g_2^{-1}$ is in the kernel $\ker(f)=\{e\}$, and hence $g_1g_2^{-1}=e$.
This implies that we have $g_1=g_2$ and $f$ is injective. Add to solve later

### 2 Responses

1. Darie says:

Thanks a lot, very nicely explained and laid out ! Keep up the great work !

• Yu says:

Dear Darie, Let $\R^{\times}=\R\setminus \{0\}$ be the multiplicative group of real numbers. Let $\C^{\times}=\C\setminus \{0\}$ be the multiplicative group of complex numbers....